Solve for r r^2+3r-8=0

$r^{2} + 3 r - 8 = 0$

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 1$ , $b = 3$ , and $c = - 8$ into the quadratic formula and solve for $r$ .

$\frac{- 3 \pm \sqrt{3^{2} - 4 \cdot (1 \cdot - 8)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Raise $3$ to the power of $2$ .

$r = \frac{- 3 \pm \sqrt{9 - 4 \cdot 1 \cdot - 8}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$r = \frac{- 3 \pm \sqrt{9 - 4 \cdot - 8}}{2 \cdot 1}$

Multiply $- 4$ by $- 8$ .

$r = \frac{- 3 \pm \sqrt{9 + 32}}{2 \cdot 1}$

Add $9$ and $32$ .

$r = \frac{- 3 \pm \sqrt{41}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$r = \frac{- 3 \pm \sqrt{41}}{2}$

The final answer is the combination of both solutions.

$r = - \frac{3 - \sqrt{41}}{2}, - \frac{3 + \sqrt{41}}{2}$

The result can be shown in multiple forms.

Exact Form:

$r = - \frac{3 - \sqrt{41}}{2}, - \frac{3 + \sqrt{41}}{2}$

Decimal Form:

$r = 1.70156211 \dots, - 4.70156211 \dots$